On 4/1/2014 7:40 AM, Bruno Marchal wrote:
BTW, are you OK in the math thread? Are you OK, like Liz apparently, that the Kripke
frame (W,R) respects A -> []<>A iff R is symmetrical?
Should I give the proof of the fact that the Kripke frame (W,R) respects []A -> [][]A
iff R is a transitive?
Bruno
Here's the ones I've done so far. One more to go. Hold off on that proof (or put a
warning in the subject line so I can avoid reading it).
Brent
> *******************
> Show that
>
> (W, R) respects []A -> A if and only if R is reflexive,
R is reflexive implies (alpha R alpha) for all alpha. []A in alpha implies A is true in
all beta where (alpha R beta), which includes the case beta=alpha. So R is reflexive
implies (W,R) respects []A->A.
Assume R is not reflexive. Then there exists at least one world beta such that (alpha R
beta) and ~(beta R alpha). Consider a valuation such that p=f in alpha and p=t in all
beta. Then []p is true in alpha but p is false so []A->A is false in alpha for some A. R
not reflexive implies []A->A is not respected for all alpha and all valuations.
> (W, R) respects []A -> [][]A if and only R is transitive,
R is transitive means that for all beta such that (alpha R beta) and all gamma such that
(beta R gamma), (alpha R gamma). So every []A implies A=t in all beta and also A=t in all
gamma. But A=t in all gamma means []A is true in beta, which in turn means [][]A is true
in alpha. So R is transitive implies (W,R) respects []A->[][]A.
Suppose R is not transitive, so for all beta (alpha R beta) and there are some gamma such
that [(beta R gamma) and ~(alpha R gamma)]. Let A=t in beta, A=f in gamma. Then []A is
true in alpha but []A isn't true in beta, so [][]A isn't true in alpha. So (W, R)
respects []A -> [][]A implies R is transitive.
> (W, R) respects A -> []<>A if and only R is symmetrical,
R symmetrical means that if (alpha R beta) then (beta R alpha). Suppose A is true in
alpha; then <>A is true in beta (by symmetry of R) and this holds for all alpha and beta
so []<>A in alpha.
Suppose R is not symmetrical, so there is a pair of worlds (alpha R beta) and ~(beta R
alpha). So consider V such that A=t in alpha and A=f in all worlds gamma such that (beta
R gamma) then ~<>A in beta. So it would be false that []<>A in alpha.
> (W,R) respects []A -> <>A if and only if R is ideal,
R is ideal, means that for every alpha there is a beta such that (alpha R beta). Suppose
[]A is true in alpha, then A must be true in every world beta (alpha R beta) and there is
a least on such beta, so <>A is true in alpha.
Suppose R is not ideal, then there is a cul-de-sac alpha. For alpha []A is vacously true
for all A, but <>A is false so []A-><>A is false.
> (W, R) respects <>A -> ~[]<>A if and only if R is realist.
R is realist means that for every world alpha there is a world beta such that (alpha R
beta) and beta is cul-de-sac. Suppose A is true in beta, then <>A is true in alpha but
<>A=f in beta so []<>A cannot be true in alpha. Hence <>A->~[]<>A in alpha where alpha is
any non cul-de-sac world. Then consider a cul-de-sac world like beta; <>A is always false
in beta so <>A->X is true in beta for any X, including ~[]<>A.
***********************
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